Nuclei - Result Question 90

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92. What is the respective number of $\alpha$ and $\beta$-particles emitted in the following radioactive decay ${ }^{200} X _{90} \to{ }^{168} Y _{80}$ ?

======= ####92. What is the respective number of $\alpha$ and $\beta$-particles emitted in the following radioactive decay ${ }^{200} X _{90} \to{ }^{168} Y _{80}$ ?

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/nuclei/nuclei—result-question-90.md (a) 6 and 8

(b) 6 and 6

(c) 8 and 8

(d) 8 and 6

[1995]

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Answer:

Correct Answer: 92. (d)

Solution:

  1. (d) ${ }^{200} X _{90} \longrightarrow{ }^{168} Y _{80}$. We know that

${ }^{200} X _{90} \longrightarrow n_2 He^{4}+m _{-1} \beta^{0}+{ }^{168} Y _{80}$.

Therefore, in this process, $200=4 n+168$ or $n=\frac{200-168}{4}=8$.

Also, $90=2 n-m+80$

or, $m=2 n+80-90=(2 \times 8+80-90)=6$.

Thus, respective number of $\alpha$ and $\beta$-particles will be 8 and 6 .